What are the solutions to the equation r^2-12r+45=0 ?

The solutions to the equation r^2-12r+45=0 are r=6+3i,r=6-3i

Find the zeros of r^2-12r+45=0

The zeros of r^2-12r+45=0 are r=6+3i,r=6-3i

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r^{2} - 12 r + 45 = 0

r = 6 + 3 i, r = 6 - 3 i

Solution steps

r^{2} - 12 r + 45 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 45}{2 \cdot 1}

Simplify

(- 12)^{2} - 4 \cdot 1 \cdot 45 : 6 i

r_{1, 2} = \frac{- ( - 12 ) \pm 6 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 12 ) + 6 i}{2 \cdot 1}, r_{2} = \frac{- ( - 12 ) - 6 i}{2 \cdot 1}

r = \frac{- ( - 12 ) + 6 i}{2 \cdot 1} : 6 + 3 i

r = \frac{- ( - 12 ) - 6 i}{2 \cdot 1} : 6 - 3 i

The solutions to the quadratic equation are:

r = 6 + 3 i, r = 6 - 3 i

\frac{49}{c} \leq 14

0 = (2 y - 1) (8 - y)

216 x^{3} - 27

\frac{( x - 7 )}{( 4 x - 15 )} > 8

- 16 = \frac{8 x}{5}

What are the solutions to the equation r^2-12r+45=0 ?
The solutions to the equation r^2-12r+45=0 are r=6+3i,r=6-3i
Find the zeros of r^2-12r+45=0
The zeros of r^2-12r+45=0 are r=6+3i,r=6-3i